binomial theorem
binomial theorem
A. Binomial Theorem
The formula by which any positive integral power of a binomial expression can be expanded in the form of a series is known as Binomial Theorem. If x, y Î R and n Î N, then
(x + y)n = nC0xn + nC1xn _ 1y + nC2xn _ 2 y2 + ........+ nCrxn_ryr + ......+ nCnyn = .
This theorem can be proved by induction.
Observations :
(a) The number of terms in the expansion is (n + 1) i.e. one more than the index.
(b) The sum of the indices of x & y in each term is n.
(c) The binomial coefficients of the terms (nC0, nC1.........) equidistant from the beginning and the end are equal.
Ex.1 The value of is
Sol. The numerator is of the form a3 + b3 + 3ab (a + b) = (a + b)3 where a = 18, and b = 7
Nr = (18 + 7)3 = (25)3. Denominator can be written as
= (3 + 2)6 = 56 = (25)3
B. IMPORTANT TERMS IN THE BINOMIAL EXPANSION ARE
(a) General term : The general term or the (r + 1)th term in the expansion of (x + y)n is given by
Tr+1 = nCrxn_r. yr
Ex.2 Find : (a) The coefficient of x7 in the expansion of
(b) The coefficient of x_7 in the expansion of
Also, find the relation between a and b, so that these coefficients are equal.
Sol. (a) In the expansion of , the general terms is Tr + 1 =
putting 22 _ 3r = 7 Þ 3r = 15 Þ r = 5 T6 =
Hence the coefficient of x7 in is 11C5 a6b_5.
(b) In the expansion of , general terms is Tr + 1 =
putting 11 _ 3r = _ 7 Þ 3r = 18 Þ r = 6
Hence the coefficient of x_7 in is 11C6a5b_6
Also given coefficient of x7 in = coefficient of x_7 in
Þ Þ ab = 1 ( 11C5 = 11C6). Which is a required relation between a and b.
Ex.3 Find the number of rational terms in the expansion of (91/4 + 81/6)1000.
Sol. The general term in the expansion of (91/4 + 81/6)1000 is Tr + 1 =
The above term will be rational if exponent of 3 and 2 are integers. i.e. and must be integers
The possible set of values of r is {0, 2, 4, ..........1000}. Hence, number of rational terms is 501
(b) Middle term : The middle term(s) in the expansion of (x + y)n is (are)
(i) If n is even, there is only one middle term which is given by T(n + 2)/2 = nCn/2. xn/2. yn/2
(ii) If n is odd, there are two middle terms which are &
Ex.4 Find the middle term in the expansion of
Sol. The number of terms in the expansion of is 10 (even). So there are two middle terms.
i.e. th and th two middle terms. They are given by T5 and T6
T5 =T4 + 1 =
and
(c) Term independent of x : Term independent of x contains no x ; Hence find the value of r for which the exponent of x is zero.
Ex.5 The term independent of x in is
Sol. General term in the expansion is
For constant term, Þ r= which is not an integer. Therefore, there will be no constant term.
(d) Numerically greatest term : To find the greatest term in the expansion of (x + a)n.
We have (x + a)n = xn ; therefore, since xn multiplies every term in , it will be sufficient to find the greatest term in this later expansion. Let the Tr and Tr + 1 be the rth and (r +1)th terms in the expansion of then . Let numerically, Tr + 1 be the greatest term in the above expansion. Then Tr + 1 ³ Tr Þ Þ Þ
Substituting values of n and x, we get r £ m + f or r £ m where m is a positive integer and f is fraction such that 0 < f < 1. In the first case Tm + 1 is the greatest term, while in the second case Tm and Tm + 1 are the greatest terms and both are equal.
Ex.6 Find numerically the greatest term in the expansion of (3 _ 5x)11 when x = 1/5
Sol. Since (3 _ 5x)11 = 311 . Now in the expansion of ,
we have =
Þ Þ 4r £ 12 Þ r £ 3 r = 2, 3
so, the greatest terms are T2 + 1 and T3+1. Greatest term (when r = 2)
and greatest term (when r = 3) =
From above we say that the value of both greatest terms are equal.
C. If , where I & n are positive integers, n being odd and 0 < f < 1, then (I + f). f = Kn where A _ B2 = K > 0 & . If n is an even integer, then (I + f) (1 _ f) = kn
Ex.7 If = [N] + F and F = N _ [N] ; where [*] denotes greatest integer, then NF is equal to
Sol. Since = [N] + F. Let us assume that f = ; where 0 £ f < 1.
Now, [N] + F _ f = _
Þ [N] + F _ f = even integer.
Now 0 < F < 1 and 0 < f < 1 so _1 < F _ f < 1 and F _ f is an integer so it can only be zero
Thus NF = = 202n + 1.
D. Some Results on Binomial Coefficients
(a) C0 + C1 + C2 + ............+ Cn = 2n
(b) C0 + C2 + C4 + ............= C1 + C3 + C5 + .......... = 2n _ 1
(c) C02 + C12 + C22 + ............+ Cn2 = 2nCn =
(d) C0. Cr + C1. Cr + 1 + C2. Cr + 2 +............+Cn _ r Cn =
Remember : (2n) ! = 2n. n! [1.3.5........(2n _ 1)]
Ex.8 If (1 + x)n = C0 + C1x + C2x2 +.................+ Cnxn then show that the sum of the products of the Ci¢s taken two at a time represents by : is equal to
Sol. Since (C0 + C1 + C2 + ........+Cn _ 1 + Cn)2 .......+
Þ . Hence
Ex.9 If (1 + x)n = C0 + C1x + C2x2 + .........+Cnxn then prove that
Sol. L.H.S = (C0 + C1)2 + (C0 + C2)2 + ....+ (C0 + Cn)2 + (C1 + C2)2 + (C1 + C3)2 + ....+ (C1 + Cn)2
+ (C2 + C3)2 + (C2 + C4)2 + ........+ (C2 + Cn)2 + .........+ (Cn_1 + Cn)2
= n (C02 + C12 + C22 +..........+Cn2) + {from Ex. 8}
= n. 2nCn + 22n _ 2nCn = (n _ 1). 2nCn + 22n = R.H.S.
E. Binomial theorem for negative or fractional indices
If n Î Q, then (1 + x)n = 1 + nx + + +........... ¥ provided |x| < 1.
Note :
(i) When the index n is a positive integer the number of terms in the expansion of (1 + x)n is finite i.e. (n + 1) & the coefficient of successive terms are : nC0, nC1, nC2, .........., nCn
(ii) When the index is other than a positive integer such as negative integer or fraction, the number of terms in the expansion of (1 + x)n is infinite and the symbol nCr cannot be used to denote the coefficient of the general term.
(iii) Following expansion should be remembered (|x| < 1)
(a) (1 + x)_1 = 1 _ x + x2 _ x3 + x4 _ .........¥
(b) (1 _ x)_1 = 1 + x + x2 + x3 + x4 + .........¥
(c) (1 + x)_2 = 1 _2x + 3x2 _ 4x3 + .........¥
(d) (1 _ x)_2 = 1 + 2x + 3x2 + 4x3 + .........¥
(iv) The expansions in ascending powers of x are only valid if x is 'small'. If x is large i.e. |x| > 1 then we may find it convenient to expand in powers of 1/x, which then will be small.
F. Approximations
If x < 1, the terms of the above expansion go on decreasing and if x be very small, a stage may be reached when we may neglect the terms containing higher powers of x in the expansion. Thus, if x be so small that its squares and higher powers may be neglected then (1 + x)n = 1 + nx, approximately,
This is an approximate value of (1 + x)n
Ex.10 If x is so small such that its square and higher powers may be neglected then find the approximate value of
Sol.
= 1 _ _ = 1 _
Ex.11 The value of cube root of 1001 upto five decimal places is
Sol.
= 10 {1 + 0.0003333 _ 0.00000011 + ......} = 10.00333
Ex.12 The sum of is
Sol. Comparing with 1 + nx + + ...... Þ nx = 1/4 ...(i)
& or Þ Þ Þ ...(ii) {by (i)}
putting the value of x in (i) Þ Þ n =
sum of series = (1 + x)n = (1 _ 1/2)_1/2 = (1/2)_1/2 =
G. Exponential series
(a) e is an irrational number lying between 2.7 & 2.8. Its value correct upto 10 places of decimal is 2.7182818284.
(b) Logarithms to the base 'e' are known as the Napierian system, so named after Napier, their inventor. They are also called Natural Logarithm.
(c) where x may be any real or complex number &
(d) where a > 0
(e)
H. Logarithmic series
(a) ln (1 + x) = x _ where _ 1 < x £ 1
(b) ln (1 _ x) = where _ 1 £ x < 1
Remember : (i) (ii) eln x = x (iii) ln2 = 0.693 (iv) ln 10 = 2.303
1. What is the Binomial Theorem? |
2. What is the importance of the Binomial Theorem in mathematics? |
3. What are the applications of the Binomial Theorem in real-life scenarios? |
4. How can one use the Binomial Theorem to expand binomial expressions? |
5. What is the relationship between the Binomial Theorem and Pascal's Triangle? |
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